Question: Evaluate $\left\lceil3\left(6-\frac12\right)\right\rceil$.
Solution: Firstly, $3\left(6-\frac12\right)=18-1-\frac12=17-\frac12$.  Because $0\le\frac12<1$, we have $\left\lceil17-\frac12\right\rceil=\boxed{17}$.